Answer to Question #209805 in Analytic Geometry for Faith

Question

1.let u=<-2,1,-2>,v=<-3,2,-1> and w=<1,3,5>.compute:

1.1.U*W.

1.2. U*(W*V) and (U*W)*V.

2.find a point- normal form of the equation of the plane passing through p=(1,2,-3) and having n=<2,-1,2> as a normal.

Answer 1

u = -2i + j - 2k

v = -3i + 2j - k

w = i + 3j + 5k

1.1 u*w = (-2i + j - 2k) * (-3i + 2j - k)

u * w = $\begin{vmatrix} i & j &k\\ -2 & 1& -2\\ 1&3&5\\ \end{vmatrix}$ = 11i + 8j -7k

u*w = 11i + 8j -7k ............................equation(1)

1.2 u*(w*v) = ( u.v ) w - ( u.w) v

u*(w*v) = [ (-2i + j - 2k ) . (-3i + 2j - k) ] w - [( -2i + j - 2k).( i + 3j + 5k)] v

u*(w*v) = 10w + 9 v

u*(w*v) = -17i + 38j + 49k

(u*w)*v

from equation(1) we have u*w = 11i + 8j -7k

So, (u*w)*v = ( 11i + 8j -7k ) * (-3i + 2j - k)

(u*w)*v = $\begin{vmatrix} i&j&k\\ 11 & 8 & -7 \\ -3 & 2 & -1\\ \end{vmatrix}$ = 6i + 32j + 46k

(u*w)*v = 6i + 32j + 46k

2) If a plane passes through point P (x₁,y₁,z₁) and has a normal n = (a, b, c) then the point - normal form of the plane is given by

a(x - x₁) + b(y - y₁) + c(z - z₁) = 0

Here, P = (1,2,-3) and n= 2i - j + 2k

So, the equation of plane in point - normal form is

2(x - 1) - (y - 2) + 2(z +3) = 0

2x - y + 2z + 8 = 0